What inequality does the units digit of a two digit number less than $40$ satisfy if the sum of the units digit and tens digit equals $8$?

Question

For a two digit number , the sum of the "units digit" and "tens digit" is $8$ and the two-digit number is less than $40$.

Denoting that the "units digit" of the two-digit number to be $x$, form an inequality and solve for $x$.

I'm not sure how to find the "tens digit" as I have problems understanding the question . Thanks for helping ..

Answer 1

You have two things here... let $x$ be the units digit and let $w$ be the tens digit: you know that

$$w+x=8$$ $$10w+x<40$$

Thus

$$9w+w+x<40\Rightarrow 9w+8<40\Rightarrow 9w<32\Rightarrow w<3.6$$

So $w$ can be $1,2,3$. That gives possibilities....

Answer 2

Let $t$ denote the tens digit; let $x$ denote the units digit. Since the sum of the tens digit and the units digit is $8$, $$t + x = 8$$ Since the two-digit number is less than $40$, $1 \leq t \leq 3$. Since $$x = 8 - t$$ you can determine the possible values of $x$.